Denote by ##p## the pressure exerted by the gas and by ##V## the volume below the piston. The gas undergoes a quasi-static adiabatic change so at any given instant during the piston’s oscillation the relation ##p V^{\gamma} = \text{const.}## holds so this then implies that ##(p(0) + \delta p)(V_0 + \delta V)^{\gamma} = p(0) V^{\gamma}_0## i.e. ##(p(0) + \delta p)(1 + \delta V/V_0)^{\gamma} = p(0)##. Now the oscillations about mechanical equilibrium are small so at any given instant during the oscillation, ##\delta V \ll V_0, \delta p \ll p(0)## hence $$(p(0) + \delta p)(1 + \delta V/V_0)^{\gamma} \\= (p(0) + \delta p)(1 + \gamma \delta V/V_0 + O((\delta V/V_0)^2) ) \\= p(0) + \delta p + \gamma p(0) \delta V/V_0 + O((\delta V/V_0)^2) = p(0)$$ therefore $$\gamma = -\frac{V_0}{p(0)}\frac{\delta p}{\delta V} \\= -\frac{V_0}{p(0)A}\frac{\delta (pA)}{\delta (A y)} \\= -\frac{V_0}{(mgA + p_0 A^2)}\frac{\delta F}{\delta y}$$ where ##F## is the force exerted by the gas on the piston at any given instant during the oscillation, so that ##\delta F = F – F(0)##, and ##\delta y## is the displacement from the equilibrium position at that given instant. We also have that ##F – p_0 A – mg = F_{\text{piston}} = -k \delta y = -\omega^2 m \delta y## where I have used the fact that the piston undergoes simple harmonic motion (there is no damping since the gas loses no heat during the oscillations). Then ##\delta F = (-\omega^2 m\delta y + p_0A + mg) – (p_0 A + mg) = -\omega^2 m \delta y## so ##\gamma = \frac{\omega^2 m V_0}{(mgA + p_0 A^2)} = \frac{4\pi^2 \nu^2 m V_0}{(mgA + p_0 A^2)}## where I have used the fact that ##\omega = 2\pi \nu## for the angular frequency. This is what the book lists as the correct expression for ##\gamma## but I just wanted to make sure my solution was valid, particularly the steps taken within this very paragraph. Thanks in advance!